Supremum, infimum, maximum and minimum

Hi tsr,

A=(2^n)/((2^n)+1)cos((npi)/2) such that n is an element of natural numbers. Which of the following statements is true? a)modulus of minA=modulus of maxA b)supA=infA c)modulus of supA=modulus of infA d)modulus of minA=modulus of supA

I got the max to be 1, minA to be -1, supA to be 1 and infA to be -1. So according to my answers modulus of all the above answers except b) make sense but there is only one correct answer so which one is right and how do i get to it?

Also i worked out this answer from trial and error, is there a quicker way of doing this since this was a non-calculator question.

Could you clarify - do you mean $A = \dfrac{2^n}{2^n+1} \cos(\dfrac{n \pi}{2})$?

Yes

So for n odd, $\cos(\frac{n \pi}{2}) = 0$; for n even, $\cos(\frac{n \pi}{2}) = (-1)^{n/2}$. Clearly the sequence increases in modulus with $n$, because the fraction term tends to 1, and 1 is an upper bound on the modulus of $A$. Moreover, it is the supremum of the modulus. However, it is not attained for any $n$, so the maximum does not exist.

The inf is -1, similarly, and the minimum doesn't exist.

But when e.g n=80, I do get A to be 1 and when n is 82 I get A to be -1.

How can you possibly get A to be 1? Are you doing some sort of estimation with a calculator or something?

You'll never get $\dfrac{2^n}{2^n + 1} = 1$ for any finite n.

How can you possibly get A to be 1? Are you doing some sort of estimation with a calculator or something?

You'll never get $\dfrac{2^n}{2^n + 1} = 1$ for any finite n.

I just realised my mistake, yes I did use a calculator and I didn't know that a calculator would round off the answer, I thought even the calculator would not give us 1 or -1 for any value of n, thanks for pointing that out. I know it was a non-calc question but I wasn't able to work it out without the calc, turns I out I couldn't do it with the calc as well!

It returns an answer correct to 16dp or whatever. Once the value of the expression is greater than 0.9999999999999999, it will return 1.000000000000000.

Can I also ask how you know that it will never give us 1? Is it because if we take the limit as n tends to infinity we get 1?

I just realised my mistake, yes I did use a calculator and I didn't know that a calculator would round off the answer, I thought even the calculator would not give us 1 or -1 for any value of n, thanks for pointing that out. I know it was a non-calc question but I wasn't able to work it out without the calc, turns I out I couldn't do it with the calc as well!

let me give one simple advice: take your calculator home and forget about it (assuming you are studying maths at any serious uni)

Well said - I haven't used a calculator in all my one-and-three-quarter years.

I can't even think of an undergrad topic where a calculator would be any use...

There was one question in a numerical analysis example sheet which asked you to simulate a five-decimal-place computer. By hand. Performing approximately twenty additions, ten multiplications and five divisions.

And your problem with that is?

I can't even think of an undergrad topic where a calculator would be any use...

You sadist!

Can I also ask how you know that it will never give us 1? Is it because if we take the limit as n tends to infinity we get 1?